Problem: A particle moving in the $xy$ -plane has velocity vector given by $v(t)=\left(3t,5t^2\right)$ for time $t\geq 0$. What is the magnitude of the displacement of the particle between time $t=1$ and $t=2$ ? Round to the nearest tenth.
Answer: To find the magnitude of the displacement of the particle, we should first find the particle's horizontal displacement $\Delta x$ and the particle's vertical displacement $\Delta y$. Then we can find the magnitude of the displacement using the distance formula: $\text{Magnitude of displacement }=\sqrt{(\Delta x)^2+(\Delta y)^2}$ The particle's horizontal displacement can be found by taking the definite integral of the horizontal component of $v(t)$ between time $t=1$ and $t=2$ : $\Delta x=\int_1^2 3t\,dt=\dfrac{9}{2}$ The particle's vertical displacement can be found by taking the definite integral of the vertical component of $v(t)$ between time $t=1$ and $t=2$ : $\Delta y=\int_1^2 5t^2\,dt=\dfrac{35}{3}$ Now we can find the magnitude of the displacement: $\begin{aligned} &\phantom{=}\sqrt{(\Delta x)^2+(\Delta y)^2} \\\\ &=\sqrt{\left(\dfrac{9}{2}\right)^2+\left(\dfrac{35}{3}\right)^2} \\\\ &\approx 12.5 \end{aligned}$ In conclusion, the magnitude of the displacement of the particle between time $t=1$ and $t=2$ is $12.5$ units.